a chemist has three different acid solutions the first solution contains 15% acid the second contains 35% acid and the third contains 80% acid she created 270 liters of a 25% acid mixture using all three solutions the number of liters of 80% solution used is 2 times the number of liters of 35% solution used How many liters of each solution was usedThe chemist used__ liters of 15% solution_ liters of 35% solutionand___liters of 80% solution

a chemist has three different acid solutions the first solution contains 15% acid the second contains 35% acid and the third contains 80% acid she created 270 liters of a 25% acid mixture using all three solutions the number of liters of 80% solution used is 2 times the number of liters of 35% solution used How many liters of each solution was used

Solution

for this case we have the following info given:

x= number of liters for the 15% acid

y= number of liters for the 35% acid

z= number of lieters for the 80% acid

We know that the final concentration is 25%

We also know that z=2y

And we can create the following equations:

0.15x + 0.35 y + 0.80z = 270*0.25 (Acid)

0.15x +0.35 y+ 0.8(2y)= 67.5

0.15x + 1.95 y = 67.5

We can solve for x and we got:

0.15x = 67.5 -1.95 y

x = (67.5 -1.95y)/0.15

0.85 x + 0.65 y + 0.2 z= 270*(1-0.25) (Water)

0.85x + 0.65y + 0.2(2y)= 202.5

0.85x + 1.05 y = 202.5

And replacing x we got:

0.85 [(67.5 -1.95y)/0.15] +1.05 y = 202.5

382.5 - 11.05 y +1.05 y =202.5

382.5 -202.5 = 10y

y= 180/10= 18 L of 35% solution

x= (67.5 -1.95*18)/0.15= 216 L of 15% solution

z =2* 18= 36L of 80% solution